Find the area of the region bounded by the li | vedclass

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(A) As shown in the figure,the line $y=3x+2$ meets the $x$-axis at $x=-2/3$. The graph lies below the $x$-axis for $x \in (-1, -2/3)$ and above the $x

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$13/3$

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(A) As shown in the figure,the line $y=3x+2$ meets the $x$-axis at $x=-2/3$. The graph lies below the $x$-axis for $x \in (-1, -2/3)$ and above the $x$-axis for $x \in (-2/3, 1)$.The required area is given by:$	ext{Area} = \left| \int_{-1}^{-2/3} (3x+2) dx ight| + \int_{-2/3}^{1} (3x+2) dx$First,evaluate the integral $\int (3x+2) dx = \frac{3x^2}{2} + 2x$.For the first part:$\left| \left[ \frac{3x^2}{2} + 2x ight]_{-1}^{-2/3} ight| = \left| \left( \frac{3(-2/3)^2}{2} + 2(-2/3) ight) - \left( \frac{3(-1)^2}{2} + 2(-1) ight) ight|$$= \left| \left( \frac{3(4/9)}{2} - 4/3 ight) - \left( 3/2 - 2 ight) ight| = \left| (2/3 - 4/3) - (-1/2) ight| = \left| -2/3 + 1/2 ight| = \left| -1/6 ight| = 1/6$.For the second part:$\left[ \frac{3x^2}{2} + 2x ight]_{-2/3}^{1} = \left( \frac{3(1)^2}{2} + 2(1) ight) - \left( \frac{3(-2/3)^2}{2} + 2(-2/3) ight)$$= (3/2 + 2) - (-1/6) = 7/2 + 1/6 = 21/6 + 1/6 = 22/6 = 11/3$.Total Area $= 1/6 + 11/3 = 1/6 + 22/6 = 23/6$. Wait,let's re-calculate: $1/6 + 22/6 = 23/6$. Let's re-check the integral: $\int_{-2/3}^{1} (3x+2) dx = [3/2 + 2] - [2/3 - 4/3] = 7/2 - (-2/3) = 7/2 + 2/3 = 21/6 + 4/6 = 25/6$. Total Area $= 1/6 + 25/6 = 26/6 = 13/3$.

Find the area of the region bounded by the line $y=3x+2$,the $x$-axis,and the ordinates $x=-1$ and $x=1$.

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